Đáp án:
$\begin{array}{l}
2A)\\
a)x = \dfrac{{ - 1}}{3} \Rightarrow \left| x \right| = - \dfrac{{ - 1}}{3} = \dfrac{1}{3}\\
A = 3{x^3} - 6{x^2} + 2\left| x \right| + 7\\
= 3.{\left( {\dfrac{{ - 1}}{3}} \right)^3} - 6.{\left( { - \dfrac{1}{3}} \right)^2} + 2.\dfrac{1}{3} + 7\\
= 3.\dfrac{1}{{{3^3}}} - 6.\dfrac{1}{{{3^2}}} + \dfrac{2}{3} + 7\\
= \dfrac{1}{9} - \dfrac{2}{3} + \dfrac{2}{3} + 7\\
= \dfrac{1}{9} + 7\\
= \dfrac{{64}}{9}\\
b)x = \dfrac{1}{4} \Rightarrow \left| x \right| = \dfrac{1}{4}\\
y = - 2 \Rightarrow \left| y \right| = 2\\
B = 4\left| x \right| - 2\left| y \right|\\
= 4.\dfrac{1}{4} - 2.2 = 1 - 4 = - 3\\
2B)\\
a)x = \dfrac{{ - 2}}{3} \Rightarrow \left| x \right| = \dfrac{2}{3}\\
C = 6{x^3} - 3{x^2} + 2\left| x \right| + 4\\
= 6.{\left( {\dfrac{{ - 2}}{3}} \right)^3} - 3.{\left( {\dfrac{{ - 2}}{3}} \right)^2} + 2.\dfrac{2}{3} + 4\\
= - 6.\dfrac{8}{{27}} - 3.\dfrac{4}{9} + \dfrac{4}{3} + 4\\
= \dfrac{{ - 18}}{9} - \dfrac{{12}}{9} + \dfrac{{12}}{9} + 4\\
= - 2 + 4\\
= 2\\
b)x = \dfrac{1}{2} \Rightarrow \left| x \right| = \dfrac{1}{2}\\
y = - 3 \Rightarrow \left| y \right| = 3\\
\Rightarrow D = 2\left| x \right| - 3\left| y \right|\\
= 2.\dfrac{1}{2} - 3.3\\
= 1 - 9 = - 8
\end{array}$
